Different ways of solving a contest problem (Part 3)

Precalculus 1 Minute

The following problem appeared on the American High School Mathematics Examination (now called the AMC 12) in 1988:

If 3 \sin \theta = \cos \theta, what is \sin \theta \cos \theta?

When I presented this problem to a group of students, I was pleasantly surprised by the amount of creativity shown when solving this problem.

Yesterday, I presented a solution using a Pythagorean identity, but I was unable to be certain if the final answer was a positive or negative without drawing a picture. Here’s a third solution that also use a Pythagorean trig identity but avoids this difficulty. Again, I begin by squaring both sides.

9 \sin^2 \theta = \cos^2 \theta

9 (1 - \cos^2 \theta) = \cos^2 \theta

9 - 9 \cos^2 \theta = \cos^2 \theta

9 = 10 \cos^2 \theta

\displaystyle \frac{9}{10} = \cos^2 \theta

\displaystyle \pm \frac{3}{\sqrt{10}} = \cos \theta

Yesterday, I used the Pythagorean identity again to find \sin \theta. Today, I’ll instead plug back into the original equation 3 \sin \theta = \cos \theta:

3 \sin \theta = \cos \theta

3 \sin \theta = \displaystyle \frac{3}{\sqrt{10}}

\sin \theta = \displaystyle \pm \frac{1}{\sqrt{10}}

Unlike the example yesterday, the signs of \sin \theta and \cos \theta must agree. That is, if \cos \theta = \displaystyle \frac{3}{\sqrt{10}}, then \sin \theta = \displaystyle \frac{1}{\sqrt{10}} must also be positive. On the other hand, if \cos \theta = \displaystyle -\frac{3}{\sqrt{10}}, then \sin \theta = \displaystyle -\frac{1}{\sqrt{10}} must also be negative.

If they’re both positive, then

\sin \theta \cos \theta = \displaystyle \left( \frac{1}{\sqrt{10}} \right) \left( \frac{3}{\sqrt{10}} \right) =\displaystyle \frac{3}{10},

and if they’re both negative, then

\sin \theta \cos \theta = \displaystyle \left( -\frac{1}{\sqrt{10}} \right) \left( -\frac{3}{\sqrt{10}} \right) = \displaystyle \frac{3}{10}.

Either way, the answer must be \displaystyle \frac{3}{10}.

This is definitely superior to the solution provided in yesterday’s post, as there’s absolutely no doubt that the product \sin \theta \cos \theta must be positive.

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Published by John Quintanilla

I'm a Professor of Mathematics and a University Distinguished Teaching Professor at the University of North Texas. For eight years, I was co-director of Teach North Texas, UNT's program for preparing secondary teachers of mathematics and science.

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One thought on “Different ways of solving a contest problem (Part 3)”

  1. Pingback: Different Ways of Solving a Contest Problem: Index | Mean Green Math

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